---
title: "Space wave over plane and actual earth"
subject: "Propagation and Antenna"
chapter: 4
key: "semester_6/propagation_and_antenna:space-wave-plane-actual-earth"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/space-wave-plane-actual-earth
site: BCT Engineering
---

# Space wave over plane and actual earth

Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum

- Asked in: 2070 Magh, 10 marks; 2070 Bhadra, 5 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/space-wave-plane-actual-earth
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4

## Question

Explain the effect of space wave propagation on the ground of plane and actual earth.

## Short answer

Space wave propagates via direct and ground-reflected waves. On plane earth, field strength decreases as 1/d² and forms lobes; antenna height helps. On actual curved earth, the radio horizon limits range, effective heights are reduced, the reflected wave is weakened by divergence, and a shadow zone appears beyond the horizon.

## Answer

**Space wave propagation** is the mode in which the signal reaches the receiver by the **direct wave** plus the **ground-reflected wave** through the troposphere. It is used at VHF, UHF and microwaves. The ground affects it in two ways: by reflection (interference of the two waves) and, for real earth, by curvature.

#### (a) Over a plane (flat) earth
```text
   Tx                          Rx
   |\ ------ direct r1 -------> |
 ht|  \                      / | hr
   |    \  reflected r2    /   |
 ==|======\______P______/======|==
   |<------------ d ---------->|
```
- Path lengths: r₁ = √(d² + (ht − hr)²), r₂ = √(d² + (ht + hr)²).
- For d ≫ ht, hr: path difference Δr = r₂ − r₁ ≈ 2ht·hr/d.
- Phase difference = (2π/λ)(2ht·hr/d) + π (reflection at grazing angle gives coefficient ≈ −1).
- Resultant field (E₀/d = field of direct wave at distance d):
```text
E = (2E₀/d) · sin(2π ht hr / λd)
For large d (small angle):  sin x ≈ x
E ≈ 4π E₀ ht hr / (λ d²)
In practical form: E = 88 √P · ht · hr / (λ d²) V/m
  (P in W, heights, λ and d in metres)
```
**Effects:**
- Field strength varies as 1/d² (not 1/d) at large distance, so it falls fast.
- Near the transmitter, maxima and minima (lobes) occur as d changes, due to interference.
- Raising either antenna increases the field directly (E ∝ ht·hr), so high masts are used.
- Horizontal and vertical polarisation behave similarly at grazing incidence over most ground.

#### (b) Over the actual (curved) earth
1. **Radio horizon:** curvature limits the line-of-sight range. With standard refraction (effective radius 4R/3):
```text
d = √(2kR ht) + √(2kR hr) ≈ 4.12 (√ht + √hr) km   (h in m)
```
2. **Effective antenna heights:** the curved ground rises between the antennas, so heights measured from the tangent plane at the reflection point are reduced:
```text
h't = ht − d1²/(2kR),   h'r = hr − d2²/(2kR)
```
   These reduced heights are used in the field formula, giving a weaker field than for plane earth.
3. **Divergence factor:** reflection from a convex surface spreads the reflected rays, so the reflected wave is weaker; the reflection coefficient is multiplied by a divergence factor D < 1.
4. **Shadow zone:** beyond the horizon, only weak diffracted fields exist; the field falls rapidly.
5. **Tropospheric refraction:** decrease of refractive index with height bends rays down and extends the horizon by about 15% (k = 4/3); abnormal conditions give super-refraction, ducting or sub-refraction.
6. **Terrain, obstacles and roughness:** hills and buildings block or diffract the wave; rough ground gives diffuse rather than mirror reflection (Rayleigh criterion); Fresnel-zone clearance (≥ 0.6 of first zone) is needed for a clean link.

**Conclusion of comparison:** flat-earth analysis fits short links; for longer links, curvature reduces effective heights, limits range to the radio horizon, and weakens the reflected wave.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For d >> ht, hr over plane earth, the path difference between reflected and direct waves is approximately:
   A. 2 ht hr / d
   B. (ht + hr)/d
   C. ht hr / 2d
   D. d / (2 ht hr)
   Answer: A. r2 - r1 = sqrt(d^2+(ht+hr)^2) - sqrt(d^2+(ht-hr)^2) is about 2 ht hr/d. (Key point: Plane earth path difference)

2. At large distance over plane earth, the field strength varies as:
   A. 1/d
   B. 1/d^3
   C. 1/d^2
   D. d
   Answer: C. E is approximately 4 pi E0 ht hr/(lambda d^2), so it falls faster than the free-space 1/d. (Key point: Distance dependence)

3. Raising the transmitting or receiving antenna (plane earth) will:
   A. Decrease the field, since the path gets longer
   B. Increase the field, since E is proportional to ht x hr
   C. Have no effect on the field
   D. Only change polarisation
   Answer: B. That is why tall masts are used for space-wave links. (Key point: Effect of antenna height)

4. On actual curved earth, compared with plane earth, the reflected wave is:
   A. Stronger, because of focusing
   B. Unchanged
   C. Absent for all distances
   D. Weaker, because of a divergence factor D < 1 and reduced effective antenna heights
   Answer: D. The convex surface spreads reflected rays and reduces effective heights, giving a weaker field. (Key point: Curved-earth effects)

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